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Question
Observe the following pattern
22 − 12 = 2 + 1
32 − 22 = 3 + 2
42 − 32 = 4 + 3
52 − 42 = 5 + 4
and find the value of
992 − 962
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Solution
From the pattern, we can say that the difference between the squares of two consecutive numbers is the sum of the numbers itself.
In a formula:
`(n+1)^2-(n)^2=(n+1)+n`
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What will be the units digit of the square of the following number?
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From the pattern, we can say that the sum of the first n positive odd numbers is equal to the square of the n-th positive number. Putting that into formula:
1 + 3 + 5 + 7 + ... n = n2, where the left hand side consists of n terms.
Observe the following pattern
22 − 12 = 2 + 1
32 − 22 = 3 + 2
42 − 32 = 4 + 3
52 − 42 = 5 + 4
and find the value of
1002 − 992
Observe the following pattern \[1 = \frac{1}{2}\left\{ 1 \times \left( 1 + 1 \right) \right\}\]
\[ 1 + 2 = \frac{1}{2}\left\{ 2 \times \left( 2 + 1 \right) \right\}\]
\[ 1 + 2 + 3 = \frac{1}{2}\left\{ 3 \times \left( 3 + 1 \right) \right\}\]
\[1 + 2 + 3 + 4 = \frac{1}{2}\left\{ 4 \times \left( 4 + 1 \right) \right\}\]and find the values of following:
31 + 32 + ... + 50
Observe the following pattern \[1^2 = \frac{1}{6}\left[ 1 \times \left( 1 + 1 \right) \times \left( 2 \times 1 + 1 \right) \right]\]
\[ 1^2 + 2^2 = \frac{1}{6}\left[ 2 \times \left( 2 + 1 \right) \times \left( 2 \times 2 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 = \frac{1}{6}\left[ 3 \times \left( 3 + 1 \right) \times \left( 2 \times 3 + 1 \right) \right]\]
\[ 1^2 + 2^2 + 3^2 + 4^2 = \frac{1}{6}\left[ 4 \times \left( 4 + 1 \right) \times \left( 2 \times 4 + 1 \right) \right]\] and find the values :
52 + 62 + 72 + 82 + 92 + 102 + 112 + 122
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Find the square of the following number:
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